As the title of the question susggests, I would like to show that

The "trivial bound is that"

\begin{align*} \lim_{Q\to\infty}\lim_{N\to\infty}\frac{1}{N\pi(Q)}\sum_{n<N}\left|\sum_{\substack{p<Q\\p|n}}p-\pi(Q)\right|&\leq \lim_{Q\to\infty}\lim_{N\to\infty}\frac{1}{N\pi(Q)}\sum_{n<N}\left(\sum_{\substack{p<Q\\p|n}}p+\pi(Q)\right)\\ &=2 \end{align*}

where the equality is obtained by noting that $\frac{1}{p}$ numbers are multiplies of $p$, so the expected value of $\sum_{\substack{p<q \\ p|n}}p$ is exactly $\sum_{p<Q}1=\pi(Q)$. Thus, we are looking only for a "$o(\cdot)$" improvement. The first thought of mine would be to note that since $\sum_{\substack{p<q \\p|n}}p$ is an additive function and so by the Turan-Kubilius inequality

$$\sum_{n<N}\left|\sum_{\substack{p<Q\\p|n}}p-\pi(Q)\right|^2\leq 4N\sum_{p<Q}p$$

The issue is, this inequality is worse than the trivial one since apply Cauchy-Shwartz we get that this inequality yields

\begin{align*} \frac{1}{N\pi(Q)}\sum_{n<N}\left|\sum_{\substack{p<Q\\p|n}}p-\pi(Q)\right|&\leq \frac{1}{\pi(Q)}\sqrt{\frac{1}{N}\sum_{n<N}\left|\sum_{\substack{p<Q\\p|n}}p-\pi(Q)\right|^2}\\ &\leq \frac{2}{\pi(Q)}\sqrt{\sum_{p<Q}p}\\ \end{align*}

where by the PNT this last term is on the order of $\sqrt{\frac{Q^2}{\log(Q)}}=\frac{Q}{\sqrt{\log(Q)}}$. This inequality would indicate to us that the sum would $\mathit{diverge}$, namely

$$\lim_{N\to\infty}\frac{1}{N\pi(Q)}\sum_{n<N}\left|\sum_{\substack{p<Q\\p|n}}p-\pi(Q)\right|=\Omega(\sqrt{\log(Q)})$$

I have proved that the sum can't go to zero too fast, and specifically that

$$\lim_{N\to\infty}\frac{1}{N\pi(Q)}\sum_{n<N}\left|\sum_{\substack{p<Q\\p|n}}p-\pi(Q)\right|=\Omega(\log(Q)^{-\epsilon})$$

for any $\epsilon>0$. It does, however, feel natural that this sum should be at least $o(1)$.